$\therefore\text{S}_1\text{P}-\text{S}_2\text{P}=\text{x}=(2\text{n}+1)\frac{\lambda}{2}$
From the figure, we get,$\Rightarrow\sqrt{\text{X}^2+(2\lambda)^2}-\text{Z}=(2\text{n}+1)\frac{\lambda}{2}$
$\Rightarrow\text{Z}^2+4\lambda^2=\text{Z}^2+(2\text{n}+1)^2\frac{\lambda^2}{4}+\text{Z}(2\text{n}+1)\lambda$
$\Rightarrow\text{Z}=\frac{4\lambda^2-(2\text{n}+1)^2\Big(\lambda^2{4}\Big)}{(2\text{n}+1)\lambda}=\frac{16\lambda^2-(2\text{n}+1)^2\lambda^2}{4(2\text{n}+1)\lambda}\ ...(1)$
Putting, $\text{n}=0\Rightarrow\text{Z}=\frac{15\lambda}{4}$$\text{n}=-1\Rightarrow\text{Z}=\frac{-15\lambda}{4}$
$\text{n}=1\Rightarrow\text{Z}=\frac{7\lambda}{12}$
$\text{n}=2\Rightarrow\text{Z}=\frac{-9\lambda}{20}$
$\therefore\text{Z}=\frac{7\lambda}{12}$ is the smallest distance for which there will be minimum intensity.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
